3 ms·
What I’m driving at is, if the loop really doesn’t terminate, it would still be safe to do that optimization because the incorrectly-optimized code would never
by iainmerrick 7y ago
What I’m driving at is, if the loop really doesn’t terminate, it would still be safe to do that optimization because the incorrectly-optimized code would never be executed.
I guess that doesn’t necessarily help in the “+=2” case, where you probably want the optimizer to do a “result += x/2”.
In general, I’d greatly prefer to work with a compiler that detected the potential infinite loop and flagged it as an error.